introductory chemistry , 3 rd edition nivaldo tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Introductory Chemistry, 3rd EditionNivaldo Tro

Solutions

2009, Prentice Hall

Solutions

• What is in a solution

• How do we make them

• How do we dilute them

• Does pressure and temperature make a difference

• How do solutions mix in our body

Tro's Introductory Chemistry, Chapter 13

2


3

Tragedy in Cameroon• Lake Nyos

Lake in Cameroon, West Africa.On August 22, 1986, 1,700 people

and 3,000 cattle died.

• Released carbon dioxide cloud.CO2 seeps in from underground

and dissolves in lake water to levels above normal saturation.

Though not toxic, CO2 is heavier than air—the people died from asphyxiation.


4

Tragedy in Cameroon:A Possible Solution

• Scientists have studied Lake Nyos and similar lakes in the region to try and keep such a tragedy from reoccurring.

• Currently, they are trying to keep the CO2 levels in the lake water from reaching the very high supersaturation levels by venting CO2 from the lake bottom with pipes.


5

Solutions

• Homogeneous mixtures.Appears to be one substance, though really

contains multiple materials.E.g., air and lake water.

• Heterogeneous mixturesAppears to have two or more substances, you

can see the individual componentsChocolate chip cookies, people in a room, your

neighborhood with houses, cars and trees


6

Solutions, Continued• Solute is the dissolved substance.

Seems to “disappear.”“Takes on the state” of the solvent.

• Solvent is the substance solute dissolves in.Does not appear to change state.

• When both solute and solvent have the same state, the solvent is the component present in the highest percentage.

• Solutions in which the solvent is water are called aqueous solutions.


7

Common Types of Solution

Solution phaseSolute phase

Solvent phase Example

Gaseous solutions Gas Gas Air (mostly N2 and O2)

Liquid solutions

Gas

Liquid

Solid

Liquid

Liquid

Liquid

Soda (CO2 in H2O)

Vodka (C2H5OH in H2O)

Seawater (NaCl in H2O)

Solid solutions Solid Solid Brass (Zn in Cu)


8

Solubility• When one substance (solute) dissolves in another

(solvent) it is said to be soluble.Salt is soluble in water.Bromine is soluble in methylene chloride.

• When one substance does not dissolve in another it is said to be insoluble.Oil is insoluble in water.

• The solubility of one substance in another depends on two factors: nature’s tendency towards mixing and the types of intermolecular attractive forces.


9

Will It Dissolve?• Chemist’s rule of thumb:

Like dissolves like• A chemical will dissolve in a

solvent if it has a similar structure to the solvent.

• When the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other.

Classifying Solvents

Solvent Class

Structural

feature

Water, H2O Polar O-H

Ethyl alcohol, C2H5OH Polar O-H

Acetone, C3H6O Polar C=O

Toluene, C7H8 Nonpolar C-C and C-H

Hexane, C6H14 Nonpolar C-C and C-H

Diethyl ether, C4H10O Nonpolar C-C, C-H, and

C-O


11

Salt Dissolving in Water


12

Practice—Decide if Each of the Following Will Be Significantly Soluble in Water.

• potassium iodide, KI

• octane, C8H18

• methanol, CH3OH

• copper, Cu

• cetyl alcohol, CH3(CH2)14CH2OH

• iron(III) sulfide, Fe2S3

• potassium iodide, KI soluble.

• octane, C8H18 insoluble.

• methanol, CH3OH soluble.

• copper, Cu insoluble.

• cetyl alcohol, CH3(CH2)14CH2OH insoluble.

• iron(III) sulfide, Fe2S3 insoluble.


15

Supersaturated Solution

A supersaturated solution has more dissolved solute thanthe solvent can hold. When disturbed, all the solute abovethe saturation level comes out of solution.


16

Adding Solute to various Solutions

Unsaturated

Saturated

Supersaturated


17

Electrolytes• Electrolytes are substances whose

aqueous solution is a conductor of electricity.

• In strong electrolytes, all the electrolyte molecules are dissociated into ions.

• In nonelectrolytes, none of the molecules are dissociated into ions.

• In weak electrolytes, a small percentage of the molecules are dissociated into ions.

18

Solubility and Temperature• The solubility of the solute in the solvent depends on the

temperature.Higher temperature = Higher solubility of solid in liquid.Lower temperature = Higher solubility of gas in liquid.


19

Solubility of Gases:Effect of Temperature

• Many gases dissolve in water.However, most have very limited solubility.

• The solubility of a gas in a liquid decreases as the temperature increases.Bubbles seen when tap water is heated (before

the water boils) are gases that are dissolved, coming out of the solution.

Opposite of solids.


20

Solubility of Gases:Effect of Pressure

• The solubility of a gas is directly proportional to its partial pressure.Henry’s law.The solubility of solid is not effected by

pressure.

• The solubility of a gas in a liquid increases as the pressure increases.


21

Solubility and Pressure• The solubility of gases in water depends on the

pressure of the gas.

• Higher pressure = higher solubility.


22

Solubility and Pressure, Continued

When soda pop is sealed, the CO2 is under pressure. Opening the container lowers the pressure, which decreasesthe solubility of CO2 and causes bubbles to form.


23

Solution Concentrations


24

Describing Solutions• Solutions have variable composition.• To describe a solution, you need to describe

both the components and their relative amounts.

• Dilute solutions have low amounts of solute per amount of solution.

• Concentrated solutions have high amounts of solute per amount of solution.


25

Concentrations—Quantitative Descriptions of Solutions

• A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution.

• Concentration = Amount of solute in a given amount of solution.Occasionally amount of solvent.


26

Solution ConcentrationMolarity

• Moles of solute per 1 liter of solution.

• Used because it describes how many molecules of solute in each liter of solution.

• If a sugar solution concentration is 2.0 M , 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar:

Molarity = moles of soluteliters of solution


27

Preparing a 1.00 M NaCl Solution

Weigh out1 mole (58.45 g)of NaCl and addit to a 1.00 Lvolumetric flask.

Step 1 Step 2

Add water todissolve theNaCl, thenadd water tothe mark.

Step 3

Swirl to mix.


28

NaCl mol 2526.0NaCl g 58.44

NaCl mol 1NaCl g 5.15

Example 13.3—Calculate the Molarity of a Solution Made by Dissolving 15.5 g of NaCl in 1.50 L of Solution

The unit is correct, the magnitude is reasonable.Check:

Solve:

M = mol/L, 1 mol NaCl = 58.44 g

Solution Map:

Relationships:

15.5 g NaCl, 1.50 L solution

M

Given:

Find:

g NaCl mol NaCl

L solution M

L

molM

M 0.177 ML 1.50

NaCl mol 250.26M


31

Practice—Find the Molarity of All Ions in the Given Solutions of Strong Electrolytes.

• 0.25 M MgBr2(aq).

• 0.33 M Na2CO3(aq).

• 0.0750 M Fe2(SO4)3(aq).


32

Practice—Find the Molarity of All Ions in the Given Solutions of Strong Electrolytes,

Continued.

• MgBr2(aq) → Mg2+(aq) + 2 Br-(aq)

0.25 M 0.25 M 0.50 M

• Na2CO3(aq) → 2 Na+(aq) + CO32-(aq)

0.33 M 0.66 M 0.33 M

• Fe2(SO4)3(aq) → 2 Fe3+(aq) + 3 SO42-(aq)

0.0750 M 0.150 M 0.225 M


33

Dilution• Dilution is adding extra solvent to decrease the

concentration of a solution.

• The amount of solute stays the same, but the concentration decreases.

• Dilution Formula:Concstart solnx Volstart soln = Concfinal solnx Volfinal sol • Concentrations and volumes can be most units as

long as they are consistent. M1V1=M2V2


34

Example—What Volume of 12.0 M KCl Is Needed to Make 5.00 L of 1.50 M KCl Solution?

Given:

Initial solution Final solution

Concentration 12.0 M 1.50 M

Volume ? L 5.00 L

Find: L of initial KCl

Equation: (conc1)∙(vol1) = (conc2)∙(vol2)

L 625.0vol

M 12.0

L 5.00M 1.50vol

conc

volconcvol

1

1

1

221

Rearrange and apply equation:


35

Making a Solution by DilutionM1 x V1 = M2 x V2

M1 = 12.0 M V1 = ? LM2 = 1.50 M V2 = 5.00 L

L 6250

M 12.0L 005M 1.50

V

MVM

V

VMVM

1

1

221

2211

..

Dilute 0.625 L of 12.0 M solution to 5.00 L.


36

Example—Dilution Problems

• What is the concentration of a solution made by diluting 15 mL of 5.0% sugar to 135 mL?

• How would you prepare 200 mL of 0.25 M NaCl solution from a 2.0 M solution?

(5.0%)(15 mL) = M2 x (135 mL)M2 = 0.55%

(2.0 M) x V1 = (0.25 M)(200 mL)V1 = 25 mL

Dilute 25 mL of 2.0 M NaCl solution to 200 mL.

M1 = 5.0 % M2 = ? %V1 = 15 mL V2 = 135 mL

M1 = 2.0 M M2 = 0.25 MV1 = ? mL V2 = 200 mL


37

Practice—Determine the Concentration of the Following Solutions.

• Made by diluting 125 mL of 0.80 M HCl to 500 mL.

• Made by adding 200 mL of water to 800 mL of 400 ppm.


38

Practice—Determine the Concentration of the Following Solutions, Continued.

• Made by diluting 125 mL of 0.80 M HCl to 500 mL.

• Made by adding 200 mL of water to 800 mL of 400 ppm.

(0.80 M)(125 mL) = M2 x (500 mL)M2 = 0.20 M

(400 PPM)(800 mL) = M2 x (1000 mL)M2 = 320 PPM

M1 = 0.80 M M2 = ? MV1 = 125 mLV2 = 500 mL

M1 = 400 ppm M2 = ? ppmV1 = 800 mLV2 = 200 + 800 mL


39

Example—To What Volume Should You Dilute 0.200 L of 15.0 M NaOH to Make 3.00 M NaOH?

Since the solution is diluted by a factor of 5, the volume should increase by a

factor of 5, and it does.

M1V1 = M2V2

V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M

V2, L

Check:• Check.

Solve:• Follow the solution map to Solve the problem.

Solution Map:

Relationships:

• Strategize.

Given:Find:

• Sort information.

22

11 VM

VM

L 1.00

Lmol

3.00

L 200.0L

mol15.0

V1, M1, M2 V2


40

Practice Question 1—How Would You Prepare 400 mL of a 4.0% Solution From a 12% Solution?


41

Practice Question 1—How Would You Prepare 400 ML of a 4.0% Solution From a 12%

Solution?, Continued

(12%) x V1 = (4.0%)(400 mL)V1 = 133 mL

Dilute 133 mL of 12% solution to 400 mL.

M1 = 12 % M2 = 4.0 %V1 = ? mL V2 = 400 mL


42

Practice Question 2—How Would You Prepare 250 mL of a 3.0% Solution From a 7.5% Solution?

Practice Question 2—How Would You Prepare 250 ML of a 3.0% Solution From a 7.5% Solution?,

Continued(7.5%) x V1 = (3.0%)(250 mL)

V1 = 100 mL

Dilute 100 mL of 7.5% solution to 250 mL.

M1 = 7.5 % M2 = 3.0 %V1 = ? mL V2 = 250 mL


44

Why Do We Do That?• We spread salt on icy roads and

walkways to melt the ice.• We add antifreeze to car radiators

to prevent the water from boiling or freezing.Antifreeze is mainly ethylene

glycol.

• When we add solutes to water, it changes the freezing point and boiling point of the water.


45

Solution ConcentrationMolality, m

• Moles of solute per 1 kilogram of solvent.Defined in terms of amount of solvent, not

solution. Does not vary with temperature.

Because based on masses, not volumes.

Mass of solution = mass of solute + mass of solvent.Mass of solution = volume of solution x density.

solvent of kg

solute of mole molality


46

mol 1727.0OHC g 62.07

OHC mol 1OHC g 2.17

262

262262

Example 13.8—What Is the Molality of a Solution Prepared by Mixing 17.2 g of C2H6O2 with

0.500 kg of H2O?

The unit is correct, the magnitude is reasonable.Check:

Solve:

m = mol/kg, 1 mol C2H6O2 = 62.07 g

Concept Plan:

Relationships:

17.2 g C2H6O2, 0.500 kg H2O

m

Given:

Find:

g C2H6O2 mol C2H6O2

kg H2Omkg

molm

M 0.554

OH kg 0.500

OHC mol 170.27

2

262

m

m


49

Drinking Seawater

Because seawater hasa higher salt concentrationthan your cells, water flowsout of your cells into theseawater to try to decreaseits salt concentration.

The net result is that, insteadof quenching your thirst,you become dehydrated.


50

Osmotic Pressure

Solvent flows through a semipermeable membrane to make thesolution concentration equal on both sides of the membrane. The pressure required to stop this process is osmotic pressure.


51

Hemolysis and Crenation

Normal red bloodcell in an isotonic

Solution.

Red blood cell ina hypotonic

solution.Water flows into

the cell, eventually causing

the cell to burst.

Red blood cell inhypertonic

solution. Water flows out

of the cell,eventually causingthe cell to distort

and shrink.

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